HW 5 solutions

9. Find two linearly independent solutions of
t 2 y ′′ − 2y = 0
of the form y = t r and using these find the general solution of
t 2 y ′′ − 2y = t 2 .
Solution The nonhomogeneous equation above is Euler’s equation (see problem 10 on pg
140 from hw 4) since it is of the form
t 2 y ′′ + αty ′ + βy = 0
where we have α = 0 and β = −2 in this case. In problem 10 you showed that y = t r would
be a solution if r was a root of
so in this case we must find the roots of
r 2 + (α − 1)r + β = 0
r 2 − r − 2 = 0
which are r1 = −1 and r2 = 2. Thus we have solutions
y1 = 1
t
and
y2 = t 2
which are clearly linearly independent. A particular solution is given by
ψ = u1y1 + u2y2
and to find u1 and u2 we will need to compute the Wronskian:
W [y1, y2] = 1 −1
(2t) −
t t2 t2 = 3.
We also need to know the RHS g(t), and here it is important to first put our equation in the
form
y ′′ + py ′ + qy = g(t)
so after doing so we in fact see that g(t) = 1. Then applying the formulas for u1 and u2 we
get
2
−gy2 −t
u1 = dt =
W 3 dt = −t3 /9
and also
gy1
u2 = dt = 1/(3t) dt = (1/3) ln |t|.
W
Then we have the particular solution
ψ = (−t 3 /9)(1/t) + (1/3)(ln |t|)(t 2 ) = −t2
9 + t2 ln |t|
.
3
Then the general solution to the nonhomogeneous equation is
or more simply
y(x) = −t2
9 + t2 ln |t| c1 2
+ + c2t
3 t
y(x) = t2 ln |t|
3
+ c1
t + c2t 2 .